If and, then verify that

(i) It is known that $A=\left(A^{\prime}\right)^{\prime}$

Therefore, we have:

$A=\left[\begin{array}{rrr}3 & -1 & 0 \\ 4 & 2 & 1\end{array}\right]$

$B^{\prime}=\left[\begin{array}{rr}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]$

$A+B=\left[\begin{array}{rrr}3 & -1 & 0 \\ 4 & 2 & 1\end{array}\right]+\left[\begin{array}{rrr}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]=\left[\begin{array}{rrr}2 & 1 & 1 \\ 5 & 4 & 4\end{array}\right]$

$\therefore(A+B)^{\prime}=\left[\begin{array}{ll}2 & 5 \\ 1 & 4 \\ 1 & 4\end{array}\right]$

$A^{\prime}+B^{\prime}=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]+\left[\begin{array}{rr}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]=\left[\begin{array}{ll}2 & 5 \\ 1 & 4 \\ 1 & 4\end{array}\right]$

Thus, we have verified that $(A+B)^{\prime}=A^{\prime}+B^{\prime}$.

(ii)

$A-B=\left[\begin{array}{rrr}3 & -1 & 0 \\ 4 & 2 & 1\end{array}\right]-\left[\begin{array}{rrr}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]=\left[\begin{array}{rrr}4 & -3 & -1 \\ 3 & 0 & -2\end{array}\right]$

$\therefore(A-B)^{\prime}=\left[\begin{array}{rr}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]$

$A^{\prime}-B^{\prime}=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]-\left[\begin{array}{rr}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]=\left[\begin{array}{rr}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]$

Thus, we have verified that $(A-B)^{\prime}=A^{\prime}-B^{\prime}$.